Define "Overall Vibration"

[spike]

Most vibration analyzers will enable you to obtain a vibration spectrum that
will show specific amplitudes of vibration and their frequency. In addition,
they will display the overall vibration level.
What is the mathematical forumla used to determine this overall level?
If you have a vibration spectrum, can you calculate the overall level?
Thanks

 

[electricpete]

Analog overall from time waveform:

(1/N)*(sqrt(sum(Xi^^2)))
where there are N time samples Xi

Digital overall from spectrum
(sqrt(sum(Xi^^2))) (*)
where Xi are the bin magnitudes.

(*)This would be the formula using spectrum from most data collectors. If you compute the fft yourself there may be a factor of N required.

 

[vanstoja]

The overall vibration level in decibels (dB) displayed by noise analyzers is nothing more than the dB summation of every frequency point or frequency bin (for octave & 1/3 octave resolution) given by:
dB(SUM)=10*LOG10(Summation of 10^(dB/10)for each frequency or bin). A simple example is to add up 4 amplitude readings each at 60 dB. The result is 10*LOG10(4X10^6)=10*6.602059991=66.02dB. The adder of 6.02 dB applies to any and all dB level where 4 readings are summed up. For 3 equal dB readings the adder is 4.77 dB.
For narrowband spectra where dB levels can vary at every resolvable frequency, the dB summation may some sort of approximation based on using wider bands such as 1/10 octave band to divide up the frequency range. I don't know the details of how this is done.

 

[Strong]

Van,
Your equation is fine for sound power levels, by for sound pressure levels the power value should be (dB/20). Sound pressure level is 20*LOG10(P/Pref).

El-Pete,
For octave or third octave sound pressure levels: convert each band dB value to linear pressure units (usually Pascals), then add together, then take 20*LOG10(Pascal sum). If the octave values were not A-weighted, and you want overall A-weight level, then apply the A-weighting corrections for each octave or thid octave value before the dB-summation procedure.

Walt

 

[GregLocock]

Strong, in the specific cases of adding 1/3 octave or 1/1 octave bands for an overall level it is dB/10

ALWAYS

Grin

(Think about the assumptions you are making in a dB/20 situation)

Cheers

Greg Locock

 

[MikeyP]

Explain please, Greg. Why is Strong wrong for the sound pressure level case (it would also equally apply to acceleration)?

M

 

[GregLocock]

I will explain. But I'd rather people thought it through for themselves. As a hint, what are you doing when you add signals together? What is a dB actually measuring?

Cheers

Greg Locock

 

[MikeyP]

I think I know what you are hinting at, but (if I'm right in what I think you are thinking!) I think you are wrong in terms of calculating overall sound pressure levels (or acceleration or displacement or velocity).

I presume you mean that dB are representations of power or energy ratios and therefore "overall level" should be a representation of the total signal POWER. In that case the total signal power would be calculated using a log sum of pressures squared (ie dB/10). However, if overall level means (for sound pressures) the sound pressure level of the signal over all frequencies, then you add the pressures as if each band was an independent source; a log sum of pressures (ie dB/20). In this second case the answer you get would be the same as you would see on a sound pressure meter...

or am I being a numpty?

M

 

[MikeyP]

Yes, I'm being a numpty.

I said it myself. You add them as statistically independent sources. It's dB/10 for anything involving band averaged signals.

M

 

[GregLocock]

Close enough.

If you have two signals of the same frequency and level and add them together the result will fall somewere between plus 6 dB and minus infinity dB, depending on their relative phase.

By definition the octave bands cannot be at the same frequency, therefore they will add as uncorrelated signals over a sufficiently long time, therefore they will add as per the 3 dB rule.

An analyser that thinks that it can use the 6 dB rule (ie dB/20) for adding different frequencies together is telling lies. The power (which is the fundamental basis of the dB) is NOT being multiplied by 4.

And now, let the discussion commence...

Cheers

Greg Locock

 

[Tmoose]

Our old IRD 885 analyzers only display "overall" based on the entire output of the "transducer" which includes the linear range, the amplified by resonance range, plus lord knows what else. At any rate, it was a number that often far exceeded the vibration apparent even in a "too high" 600,000 cpm spectrum.

Sensible specs prescribe the frequency range expected in the "overall."

IRD DataPac offers a "spectrum" overall option.

I believe CSI offered a variety of analog and digital overall options.

 

[vanstoja]

The equation I cited in my prior post is for decibel additions regardless of whether the dBs represent Sound Pressure Level, Sound Power Level or any other parameter expressed in decibels.

 

[kaiserman]

Spike:

Wow lots of comments. I too would like to comment as well. By definition, for Sound Pressure Levels, SPL=20log(p/pref) Re 2x10-5 N/m^2. This definition can be stated another way, SPL=10log(p/pref)^2. If we remember that for this last definition, the squared pressure ratio is based on "power-related" quantities (that is, p^2), this confusion can be avoided.

By way of a simple EXAMPLE_1:
Consider the effect of adding another machine in an area where other equipment is operating. Assume that the ambient sound level due to the other equipment is Lp1=90dB and the level from the machine to be added is LP2=88dB. Estimate the combine level.
SOLUTION:
Lp=10log(p/pref)^2 -or- (p/pref)^2 = antilog(Lp/10)
Thus,
(p1/pref)^2 = antilog(90/10) = 10^(90/10) = 10 x 10^8
(p2/pref)^2 = antilog(88/10) = 10^(88/10) = 6.31 x 10^8
(ptotal/pref)^2 = (p1/pref)^2 + (p2/pref)^2
(ptotal/pref)^2 = 10 x 10^8 + 6.31 x 10^8
Lptotal = 10log(16.31 x 10^8) = 92.12 dB

The same principle applies, when attempting overall sound pressure levels from 1/3-octave bands.
Lptotal = 10log(p1^2 + p2^2 + . . . + pn^2 / pref^2)
Lptotal = 10log(10^(p1/10) + 10^(p2/10)+ . . . 10^(pn/10))

Check your calculations! Record the individual SPL 1/3-octave band levels of some known source. Then, hand calculate the overall SPL. Check the overall value as recorded by the analyzer - hopefully success!

Regarding random vibration, this is somewhat more difficult, however, similar logarithmic equations exist.
EXAMPLE_2.
If you know that your overall PSD level of a profile is say 3.315gRMS. What would be the overall PSD level of say a profile that is +3dB?
Recall,
Level = 20log(PSDnew/PSDref)
PSDnew = PSDref x antilog (Level/20)
PSDnew = PSDref x 10^(Level/20)
Thus,
PSDnew = = 3.315 x 10^(+3/20) = 4.683gRMS

I hope this helps.

Thanks,
Kaiserman

 

[krb1]

Next question for overall vibration amplitude - not level.

Suppose I record vibs at a location with a triax accelerometer. I want to calculate the overall motion's spectrum, i.e., a single spectrum combining the three orthogonal accelerations. Should I vector-sum the three time signals and calculate the spectrum, or calculate the three spectra and vector-sum them?

It seems simple enough, but it has me stumped.

 

[spike]

From Spike
Great responses when dealing with sound levels. But my original questions is related to vibration amplitudes in inch/second. How would you define overall vibration in terms of the relationship to discrete frequency amplitudes in a vibration spectrum?
Thanks again.

 

[GregLocock]

vanstoja gave the correct answer in his first post

Cheers

Greg Locock

 

[electricpete]

The original poster never said anything about db scale.
The db solution may be correct, but I believe mine is more straightforward.

Digital overall from spectrum =
(sqrt(sum(Xi^^2)))
where Xi are the bin magnitudes

Is there anyone that disagrees with that?